YES
Time: 0.053

Problem:
 Equations:
  fAC(fAC(x3,x4),x5) -> fAC(x3,fAC(x4,x5))
  fAC(x3,x4) -> fAC(x4,x3)
  fAC(x3,fAC(x4,x5)) -> fAC(fAC(x3,x4),x5)
  fAC(x4,x3) -> fAC(x3,x4)
 TRS:
  fAC(fAC(i(x),i(y1)),y2) -> fAC(i(fAC(x,y1)),y2)
  fAC(one(),x) -> x
  phi(one(),x) -> x
  fAC(i(x),x) -> one()
  i(one()) -> one()
  i(i(x)) -> x
  phi(x,phi(y1,y2)) -> phi(fAC(x,y1),y2)
  fAC(i(fAC(x,y1)),y1) -> i(x)
  i(fAC(i(y1),x)) -> fAC(i(x),y1)
  fAC(i(x),i(y1)) -> i(fAC(x,y1))

Proof:
 AC-KBO Processor:
  precedence:
   i > fAC > phi ~ one
   weight function:
    [phi](x0, x1) = 6x0 + 12x1 + 5,
    
    [one] = 10,
    
    [i](x0) = 12x0 + 10,
    
    [fAC](x0, x1) = x0 + x1
   problem:
    Equations:
     fAC(fAC(x3,x4),x5) -> fAC(x3,fAC(x4,x5))
     fAC(x3,x4) -> fAC(x4,x3)
     fAC(x3,fAC(x4,x5)) -> fAC(fAC(x3,x4),x5)
     fAC(x4,x3) -> fAC(x3,x4)
    TRS:
     
   Qed